What is the extreme F(x,y)=3x^2y-2xy^2+x+C ?

The extreme F(x,y)=3x^2y-2xy^2+x+C is Saddle(0, 1/(sqrt(2))),Saddle(0,-1/(sqrt(2)))

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extreme F(x,y)=3x^2y-2xy^2+x+C

extreme points

F (x, y) = 3 x^{2} y - 2 x y^{2} + x + C

Saddle (0, \frac{1}{2}), Saddle (0, - \frac{1}{2})

Solution steps

Find the critical points:

(0, \frac{1}{2}), (0, - \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4 x

D (x, y) = - 36 x^{2} + 24 x y - 16 y^{2}

Check the sign of at each critical point

Check critical point

(0, \frac{1}{2}) :

Saddle

Check critical point

(0, - \frac{1}{2}) :

Saddle

Saddle (0, \frac{1}{2}), Saddle (0, - \frac{1}{2})

f (x) = (2 x)^{x}

\frac{7 z + 6}{z - 7} + \frac{7 z + 3}{z - 7}

f (x) = \frac{( 4 x - 7 )}{x + 3}

\frac{2 x ^{2} - 12 x - 54}{2 x ^{2} + 13 x + 21}

f (x) = 4 x^{3} - 6 x^{2}, - 1 \leq x \leq 1

What is the extreme F(x,y)=3x^2y-2xy^2+x+C ?
The extreme F(x,y)=3x^2y-2xy^2+x+C is Saddle(0, 1/(sqrt(2))),Saddle(0,-1/(sqrt(2)))